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Math Help - Calculate Volume using the Method of Cylindrical Shells - Urgent

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    Calculate Volume using the Method of Cylindrical Shells - Urgent

    Calculate Volume using the Method of Cylindrical Shells - Urgent-calculusproblem.png

    Here are some things I have done. I already found the volume since the other part of the question said to use the method of discs/washers. It is 54.0633 cubic units (correct answer).

    I found by trial and error that (since this is an online assignment) that a= 9/sqrt(82) and c =9 .

    What do I do? Why should there be two integrals, I don't understand.

    I have tried the following.


    y = 9^x becomes x = lny/ln9

    Formula

    h =
    lny/ln9
    r = y

    So I got
    integral from 9/sqrt(82) to 9 of
    Calculate Volume using the Method of Cylindrical Shells - Urgent-capture.png dy

    But that's just one integral and perhaps even incorrect. Please show me what to do.

    Thanks



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  2. #2
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    Re: Calculate Volume using the Method of Cylindrical Shells - Urgent

    the sketch of the graph is misleading ...

    note that \frac{9}{\sqrt{82}} < 1 , so there is a very small region between the curve y = \frac{9}{\sqrt{x^2+81}} and x=1 from y = \frac{9}{\sqrt{82}} to y = 1.
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  3. #3
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    Re: Calculate Volume using the Method of Cylindrical Shells - Urgent

    To evaluate the volume of your region, you need to do

    \displaystyle \begin{align*} \int_1^9{ \pi \cdot 1^2 \, dy } - \int_1^9{ \pi \cdot \left( \log_9{y} \right)^2 \,dy } + \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot 1^2 \, dy } - \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot  \left[ \sqrt{ \left( \frac{y}{9} \right)^{ -2 } - 81 } \right]^2 \,dy } &= \pi \int_1^9{ 1 \, dy } - \frac{\pi}{\left( \ln{9} \right)^2} \int_1^9{ \left( \ln{y} \right)^2 \,dy } + \pi \int_{ \frac{9}{\sqrt{82}} }^1{ 1\, dy } - 81\pi \int_{\frac{9}{\sqrt{82}}}^1{ y^{-2} - 1 \, dy } \end{align*}

    The only one of those integrals which may cause you any trouble is the second, so do

    \displaystyle \begin{align*} \int{\left( \ln{y} \right)^2 \, dy} &= \int{ \frac{y\left( \ln{y} \right)^2}{y} \, dy } \\ &= \int{ u^2 \, e^u \, du } \textrm{ after making the substitution } u = \ln{y} \end{align*}

    which you would then evaluate using integration by parts twice. See how you go
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    Re: Calculate Volume using the Method of Cylindrical Shells - Urgent

    Is this the method of cylindrical shells? I am obliged to use cylindrical shells. Can you use r (radius), h(height)? And why will my answer be in the form of a sum of two integrals (see the question -top of image)?
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    Re: Calculate Volume using the Method of Cylindrical Shells - Urgent

    using the method of cylindrical shells ...

    V = 2\pi \int_{\frac{9}{\sqrt{82}}}^1 y \left(1 - \frac{9\sqrt{1-y^2}}{y} \right) \, dy + 2\pi \int_1^9 y \left(1 - \frac{\ln{y}}{\ln{9}}\right) \, dy
    Thanks from LightRay7
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    Re: Calculate Volume using the Method of Cylindrical Shells - Urgent

    It worked.
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